Problem: The equation of a circle $C$ is $x^2+y^2-6x-16y+48 = 0$. What is its center $(h, k)$ and its radius $r$ ?
To find the equation in standard form, complete the square. $(x^2-6x) + (y^2-16y) = -48$ $(x^2-6x+9) + (y^2-16y+64) = -48 + 9 + 64$ $(x-3)^{2} + (y-8)^{2} = 25 = 5^2$ Thus, $(h, k) = (3, 8)$ and $r = 5$.